We know that the closure of a linear subspace of a normed vector space $E$ is again a linear subspace of $E$. In particular, a hyperplane in $E$ is either closed or dense.
Consider $E =\mathbb{R}[X]$, the space of polynomials in one indeterminate with coefficients in $\mathbb{R}$ (we could later replace $\mathbb{R}$ with $\mathbb{C}$). Endow $E$ with one of the norms $\lVert P \rVert_{a < b} = \int_a^b |P(t)|dt$, with $a < b$ in $\mathbb{R}$. Further, consider the family of hyperplanes $H_{x^*} = \{P \in \mathbb{R}[X] \mid P(x^*)=0 \}$.
My questions are :
What are the closures of each of the $H_{x^*}$ with respect to each of the norms $\lVert \cdot \rVert_{a < b}$?
Again with respect to each of the norms $\lVert \cdot \rVert_{a < b}$, could we state some general statement about hyperplanes of $\mathbb{R}[X]$, of the type : "all hyperplanes of $\mathbb{R}[X]$ are dense"? In particular, are the hyperplanes $H'_{x^*} = \{ P \in \mathbb{R}[X] \mid P'(x^*)=0 \}$ dense? ($P'$ denotes the derivative of $P$)
To start answering question 1), note that $X^n - 1 \xrightarrow{\lVert \cdot \rVert _{0 < 1}} -1$ with $X^n - 1 \in H_1$ and $-1 \notin H_1$ (since $\lVert X^n \rVert _{0 < 1} = \frac{1}{n+1} \to 0$), so $H_1$ is dense with respect to $\lVert \cdot \rVert _{0 < 1}$.
Thanks for your help.
Every hyperplane $H$ passing through $0$ in a vector space $E$ determines a one-dimensional space of linear functionals on $E$, $\{ \lambda : H \subset \ker \lambda\}$. If $E$ is a topological vector space, for example a normed space, then $H$ is closed if and only if a linear functional with $\ker \lambda = H$ is continuous.
On $E = \mathbb{R}[X]$, the evaluation functionals $\delta_x \colon P \mapsto P(x)$ are discontinuous for all norms $\lVert\,\cdot\,\rVert_{a < b}$, and hence the hyperplanes $H_x$ are dense in the topology induced by any of these norms.
To see that the evaluation functionals are discontinuous, note that for every $\eta > 0$ the function $f_{x,\eta} \colon [a,b] \cup [x-1,x+1] \to\mathbb{R}$ given by
$$f_{x,\eta}(y) = \max \biggl\{ \frac{\lvert y-x\rvert}{\eta},\, 0\biggr\}$$
can be uniformly approximated by polynomials on $[a,b] \cup [x-1,x+1]$ per the Weierstraß approximation theorem. If $\lVert P - f_{x,\eta}\rVert_{C([a,b]\cup [x-1,x+1])} \leqslant \varepsilon$, then $\delta_x(P) = P(x) \geqslant 1 - \varepsilon$, but
$$\lVert P\rVert_{a < b} = \int_a^b \lvert P(t)\rvert\,dt \leqslant \varepsilon\cdot (b-a) + \int_a^b f_{x,\eta}(t)\,dt \leqslant \varepsilon(b-a) + \eta.$$
Since we can make $\eta$ and $\varepsilon$ arbitrarily small, it follows that $\delta_x$ is discontinuous.
A similar argument shows that the hyperplanes $H_x' = \{ P \in E : P'(x) = 0\}$ are also dense for all $\lVert\,\cdot\,\rVert_{a < b}$ (one approximates $f_{x,\eta}$ uniformly by polynomials so that suitable primitives of these polynomials uniformly approximate $0$).
But not all hyperplanes are dense. On every normed space, there are enough continuous linear functionals to separate points, in particular for every infinite-dimensional normed space, the space of continuous linear functionals (the topological dual) is infinite-dimensional.
For the norms $\lVert\,\cdot\,\rVert_{a < b}$, we can identify the topological dual of $(E,\lVert\,\cdot\,\rVert_{a < b})$. The restriction $P \mapsto P\lvert_{[a,b]}$ is an isometric embedding $E \hookrightarrow L^1([a,b])$ with dense image. Thus the continuous linear functionals on $(E,\lVert\,\cdot\,\rVert_{a < b})$ are precisely the restrictions of the continuous linear functionals on $L^1([a,b])$, and the dual of $L^1([a,b])$ is isometrically isomorphic to $L^\infty([a,b])$ via
$$\Phi_g \colon f \mapsto \int_a^b g(t) f(t)\,dt,\quad g \in L^\infty([a,b]),\, f \in L^1([a,b]).$$
Thus the closed hyperplanes (passing through $0$) in $(E,\lVert\,\cdot\,\rVert_{a < b})$ are precisely the hyperplanes of the form
$$H_g = \biggl\{ P \in E : \int_a^b g(t)P(t)\,dt = 0\biggr\},$$
where $g$ ranges over the bounded (Lebesgue or Borel) measurable functions on $[a,b]$.